Simplify and expand the following expression: $ \dfrac{p}{p + 6}-\dfrac{4p - 2}{p + 3} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(p + 6)(p + 3)$ Multiply the first term by $\dfrac{p + 3}{p + 3}$ $ \begin{align*} \dfrac{p}{p + 6} \times \dfrac{p + 3}{p + 3} & = \dfrac{(p)(p + 3)}{(p + 6)(p + 3)} \\ & = \dfrac{p^2 + 3p}{(p + 6)(p + 3)}\end{align*} $ Multiply the second term by $\dfrac{p + 6}{p + 6}$ $ \begin{align*} \dfrac{4p - 2}{p + 3} \times \dfrac{p + 6}{p + 6} & = \dfrac{(4p - 2)(p + 6)}{(p + 3)(p + 6)} \\ & = \dfrac{4p^2 + 22p - 12}{(p + 3)(p + 6)}\end{align*} $ Now we have: $ = \dfrac{p^2 + 3p}{(p + 6)(p + 3)} - \dfrac{4p^2 + 22p - 12}{(p + 3)(p + 6)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{p^2 + 3p - (4p^2 + 22p - 12)}{(p + 6)(p + 3)} $ $ = \dfrac{p^2 + 3p - 4p^2 - 22p + 12}{(p + 6)(p + 3)} $ $ = \dfrac{-3p^2 - 19p + 12}{(p + 6)(p + 3)}$ Expand the denominator: $ = \dfrac{-3p^2 - 19p + 12}{p^2 + 9p + 18}$